Problem: A cubeword is a special type of a crossword. When building a cubeword, you start by choosing a positive integer a: the side length of the cube. Then, you build a big cube consisting of a × a × a unit cubes. This big cube has 12 edges. Then, you discard all unit cubes that do not touch the edges of the big cube. The figure below shows the object you will get for a=6.

<image>

Finally, you assign a letter to each of the unit cubes in the object. You must get a meaningful word along each edge of the big cube. Each edge can be read in either direction, and it is sufficient if one of the two directions of reading gives a meaningful word.

The figure below shows the object for a=6 in which some unit cubes already have assigned letters. You can already read the words 'SUBMIT', 'ACCEPT' and 'TURING' along three edges of the big cube.

<image>

You are given a list of valid words. Each word from the wordlist may appear on arbitrarily many edges of a valid cubeword. Find and report the number of different cubewords that can be constructed, modulo 998,244,353.

If one cubeword can be obtained from another by rotation or mirroring, they are considered distinct.

Input

The first line contains a single integer n (1 ≤ n ≤ 100,000) – the number of words.

Then, n lines follow. Each of these lines contains one word that can appear on the edges of the big cube. The length of each word is between 3 and 10, inclusive.

It is guaranteed that all words are different.

Output

Output a single integer, the number of distinct cubewords for the given list of valid words modulo 998,244,353.

Scoring

Subtask 1 (21 points): the words consist only of letters 'a' - 'f' (lowercase)

Subtask 2 (29 points): the words consist only of letters 'a' - 'p' (lowercase)

Subtask 3 (34 points): the words consist of letters 'a' - 'p' (lowercase) and 'A' - 'P' (uppercase)

Subtask 4 (16 points): the words consist of letters 'a' - 'z' (lowercase), 'A' - 'Z' (uppercase) and digits '0' - '9'

Examples

Input


1
radar


Output


1


Input


1
robot


Output


2


Input


2
FLOW
WOLF


Output


2


Input


2
baobab
bob


Output


4097


Input


3
TURING
SUBMIT
ACCEPT


Output


162


Input


3
MAN1LA
MAN6OS
AN4NAS


Output


114

Note

In the first sample, the only possibility is for the word "radar" to be on each edge of the cube.

In the second sample, there are two cubes, which are just rotations of each other – the word "robot" is on every edge, and the difference between the two cubes is whether the lower left front corner contains 'r' or 't'.

The third sample is similar to the second one. The fact that we can read the word on each edge in both directions does not affect the answer.

In the fourth sample, there is one cube with the word "bob" on each edge. There are also 2^{12} = 4096 cubes with the word "baobab" on each edge. (For each of the 12 edges, we have two possible directions in which the word "baobab" can appear.)

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
using i32 = int;
using i64 = long long;
using u8 = unsigned char;
using u32 = unsigned;
using u64 = unsigned long long;
using f64 = double;
using f80 = long double;
using i128 = i64;
using u128 = u64;
long long power(long long a, long long b, long long p) {
  if (!b) return 1;
  long long t = power(a, b / 2, p);
  t = t * t % p;
  if (b & 1) t = t * a % p;
  return t;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long px, py;
  long long d = exgcd(b, a % b, px, py);
  x = py;
  y = px - a / b * py;
  return d;
}
template <class T>
inline void freshmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
inline void freshmax(T &a, const T &b) {
  if (a < b) a = b;
}
int inp() {
  int x = 0, f = 0;
  char ch;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') f = 1;
  for (; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  return f ? -x : x;
}
long long inp_ll() {
  long long x = 0;
  int f = 0;
  char ch;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') f = 1;
  for (; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  return f ? -x : x;
}
template <class T>
bool read(T &x) {
  x = 0;
  char ch = getchar();
  if (ch == EOF) return 0;
  for (; !isdigit(ch);) {
    ch = getchar();
    if (ch == EOF) return 0;
  }
  for (; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  return 1;
}
template <class T>
void write(T x) {
  static char s[22];
  static char *it = s + 20;
  static char *end = s + 20;
  if (!x) *--it = '0';
  while (x) {
    *--it = x % 10 + '0';
    x /= 10;
  }
  for (; it < end; ++it) putchar(*it);
}
template <class T>
void writeln(T x) {
  write(x);
  putchar('\n');
}
const int MAXN = 200010;
const int INF = 1000000000;
const int MOD = 998244353;
set<string> H[12];
int id(char c) {
  if (islower(c)) return c - 'a';
  if (isupper(c)) return c - 'A' + 26;
  return c - '0' + 52;
}
int w[63][63], F[63][63][63];
int solve(int k) {
  memset(w, 0, sizeof(w));
  memset(F, 0, sizeof(F));
  int tot = 0;
  for (auto s : H[k]) {
    int x = id(s[0]), y = id(s[k - 1]);
    w[x][y]++;
    freshmax(tot, max(x, y));
  }
  for (int a = 0; a <= tot; ++a)
    for (int b = 0; b <= tot; ++b)
      for (int c = 0; c <= tot; ++c) {
        i64 tmp = 0;
        for (int d = 0; d <= tot; ++d) tmp += (i64)w[d][a] * w[d][b] * w[d][c];
        F[a][b][c] = tmp % MOD;
      }
  i64 ret = 0;
  for (int a = 0; a <= tot; ++a)
    for (int b = 0; b <= tot; ++b)
      for (int c = 0; c <= tot; ++c) {
        i64 tmp = 0;
        for (int d = 0; d <= tot; ++d)
          tmp += (i64)F[a][b][d] * F[a][c][d] % MOD * F[b][c][d];
        ret += tmp % MOD * F[a][b][c] % MOD;
      }
  return ret % MOD;
}
int main() {
  int T;
  scanf("%d", &T);
  for (int i = 1; i <= T; ++i) {
    static char s[12];
    scanf("%s", s);
    int n = strlen(s);
    H[n].insert(s);
    reverse(s, s + n);
    H[n].insert(s);
  }
  long long ans = 0;
  for (int k = 3; k <= 10; ++k) ans += solve(k);
  ans %= MOD;
  cout << ans << endl;
  return 0;
}